The pressure ( P ) and temperature ( T ) relationship of an ideal...

Given here,

$P{T}^{2} =$constant $. . . \left(1\right)</p><p>Using ideal gas equation, </p><p>$PV = nRT$</p><p>We have,$P = $\dfrac{nRT}{V}</p><p>Putting$ in equation\left(1\right),

$P{T}^{2} =$\dfrac{nRT}{V}$\times {T}^{2} =$constant

$⇒ {T}^{3} = KV . . . \left(2\right)$

Differentiating the equation with respect to $T$,

So, $\dfrac{d}{dT}\left(KV\right) = 3{T}^{2}$

$⇒ \dfrac{KdV}{dT} = 3{T}^{2}$

$⇒ dV = \dfrac{3{T}^{2}}{K}dT$

$⇒ \gammaV = \dfrac{3{T}^{2}}{K}\left(\text{as}dV = \gamma VdT\right)$

$⇒ \gamma = \dfrac{3{T}^{2}}{KV} = \dfrac{3{T}^{2}}{{T}^{3}} = \dfrac{3}{T}$